Under certain conditions, the number of bacheria present in a colony is approximaled by $f(t)=A_{0} e^{0.026 t}$, where $t$ is in minules. if $A_{0}=2,400,000$, find the number of bacteria present at. 5 minules, 10 minuts and 60 minuestes?

Step 1 ：The problem is asking for the number of bacteria present at different time intervals. The formula given is an exponential growth formula, where \(A_{0}\) is the initial amount of bacteria, \(e\) is the base of the natural logarithm (approximately equal to 2.71828), \(0.026\) is the growth rate, and \(t\) is the time in minutes.

Step 2 ：To find the number of bacteria at a certain time, we need to substitute the given time into the formula and calculate the result.

Step 3 ：Given that \(A_{0} = 2400000\) and the time values are \(t = [5, 10, 60]\) minutes.

Step 4 ：Substitute these values into the formula \(f(t)=A_{0} e^{0.026 t}\) for each time value.

Step 5 ：The number of bacteria at 5 minutes is approximately 2,520,000, at 10 minutes is approximately 2,640,000, and at 60 minutes is approximately 4,800,000.

Step 6 ：\(\boxed{\text{Final Answer: The number of bacteria at 5 minutes is approximately 2,520,000, at 10 minutes is approximately 2,640,000, and at 60 minutes is approximately 4,800,000.}}\)